OFFSET
1,9
COMMENTS
Sum_{k=1..4} T(n,k) = A005558(n).
For n >= 1, the ratio of the numbers of right or up last steps to left or down last steps is floor((n+2)/2): floor(n/2). - Roger Ford, Oct 28 2019
FORMULA
n=1: T(1,1)=1, T(1,2)=0, T(1,3)=0, T(1,4)=0;
n>1: T(n,1) = C(n,floor(n/2))*C(n-1,floor((n-1)/2)) - C(n,floor((n-1)/2))*C(n-1,floor((n-2)/2));
T(n,2) = T(n,3) = C(n-1,floor(n/2)-1)*C(n,floor(n/2)-1)/floor(n/2);
n odd: T(n,4) = T(n,2);
n even: T(n,4) = T(n,2)*((n/2-1)/(n/2+1));
For n > 1, T(n,2) = T(n,3) = A001263(n,floor(n/2)).
EXAMPLE
k= 1 2 3 4 total
N right left up down walks
1 1 0 0 0 =1
2 1 1 1 0 =3
3 3 1 1 1 =6
4 6 6 6 2 =20
There are 6 walks of 4 steps in the octant with the last step right. T(4,1)=6 RRRR, RRLR, RLRR, RUDR, RURR, RRUR.
CROSSREFS
KEYWORD
nonn,tabf,walk,more
AUTHOR
Roger Ford, Jan 02 2018
STATUS
approved